α,β-Unsaturated carbonyls — Michael addition, 1,2 vs 1,4 addition: Visual walkthrough

Step 0 — The three symbols we will lean on

Before any chemistry, three marks on paper:

A line between two letters = a shared pair of electrons (a bond). Two lines = a double bond, four electrons shared.
δ⁺ / δ⁻ (small delta) = a partial charge. Not a full stolen electron — just "electrons hang out here a bit more" (δ⁻) or "a bit less" (δ⁺). Read δ⁺ as "electron-poor, so it wants a nucleophile."
A curly arrow ↷ = "this pair of electrons moves from the arrow's tail to its head." It always tracks electrons, never atoms.

Step 1 — One carbonyl has exactly one hungry carbon

WHAT. Look at a plain carbonyl, the $C = O$ group.

WHY. Oxygen is much greedier for electrons than carbon (it is more electronegative). So the shared electrons of $C = O$ sit closer to O. Result: O becomes δ⁻ and the carbon becomes δ⁺.

PICTURE. The red carbon in the figure is the single electron-poor spot — the only place a nucleophile can land.

Figure — α,β-Unsaturated carbonyls — Michael addition, 1,2 vs 1,4 addition

$C δ^{+} x = δ^{-} O$

$C = O$ — the double bond; electrons shared unevenly.
$δ^{-}$ under O — O hoards the electrons.
$δ^{+}$ under C — the leftover electron-poverty; this is the electrophilic carbon.

This is ordinary nucleophilic addition: one target, one outcome.

Step 2 — Bolt a C=C on, and the δ⁺ leaks downhill

WHAT. Now attach a $C = C$ right next to the carbonyl so the pattern reads $C = C - C = O$ . The two double bonds sharing that middle bond is called conjugation (see Resonance and conjugation).

WHY draw resonance? Conjugation lets electrons slide along the whole chain. To see where the δ⁺ ends up, we draw resonance structures — different pictures of the same molecule, showing where electrons can flow.

PICTURE. Follow the two red curly arrows. The $C = C$ pair swings toward the carbonyl carbon; the $C = O$ pair jumps entirely onto oxygen. Oxygen ends up with a full minus, and the far carbon (the β-carbon) ends up with a full plus.

C β = C α - C = O ⟷^{+} C β - C α = C - O^{-}

Left picture — the "normal" drawing; δ⁺ hides on the carbonyl C only.
Curly arrows — electron pairs sliding away from β, toward O.
Right picture — the consequence: $^{+} C_{β}$ (electron-poor) and $O^{-}$ (electron-rich).

So the molecule now has two δ⁺ carbons: C1 (carbonyl) and C3 (β). That is the whole reason a choice exists.

Step 3 — Naming the atoms so "1,2" and "1,4" mean something

WHAT. Line up the four atoms of the conjugated unit and number them.

WHY. The labels "1,2-addition" and "1,4-addition" count across these four atoms — nothing more mysterious.

PICTURE. Read left to right: $O (1) = C (2) - C (3) = C (4)$ . (Oxygen is atom 1.)

$1 O = 2 C - 3 C = 4 C$

Atom 1 = the oxygen.
Atom 2 = the carbonyl carbon (this is C1 in the carbon-numbering; don't confuse the two schemes).
Atom 3 = the α-carbon.
Atom 4 = the β-carbon — the far electron-poor spot from Step 2.

Step 4 — Path A: 1,2-addition (attack the near carbon)

WHAT. A nucleophile attacks atom 2. Its electrons form a new $N u - C$ bond; the $C = O$ electrons are pushed up onto O, which becomes an $O^{-}$ (an alkoxide). A proton then caps the oxygen.

WHY this path? Atom 2 is the most electron-poor point and it is out in the open (uncrowded). Attacking here is fast — the kinetic choice (see Kinetic vs thermodynamic control).

PICTURE. Red arrow: Nu's electrons swing into atom 2; the old $C = O$ π-pair folds onto O.

N u^{-} + 2 C = 1 O ⟶ N u - 2 C - 1 O^{-} H^{+} N u - 2 C - 1 O H

$N u^{-}$ — the incoming electron-rich attacker.
New $N u - C$ bond — Nu's pair became a bond at atom 2.
$O^{-}$ — the electrons had to go somewhere; they parked on oxygen.
$H^{+}$ then $O H$ — proton caps the alkoxide on workup.

Key outcome: the original $C = C$ was never touched — it survives. The product is an allylic alcohol ( $O H$ on a carbon next to a $C = C$ ).

Step 5 — Path B: 1,4-addition (attack the far carbon)

WHAT. Instead, the nucleophile attacks atom 4 (the β-carbon). Its electrons form $N u - C_{β}$ ; the $C = C$ pair shifts onto the α-carbon, which shoves the $C = O$ pair onto oxygen. We land on an enolate — a carbanion-next-to-carbonyl, drawn as $C = C - O^{-}$ .

WHY this path? The β-carbon's δ⁺ is diffuse and polarisable ("soft"), so soft nucleophiles bond there happily. And critically, it leads to a more stable end product (Step 6).

PICTURE. Red arrow: Nu attacks the far end; a domino of electron shifts runs all the way to oxygen.

N u^{-} + 4 C = 3 C - 2 C = 1 O ⟶ N u - 4 C - 3 C = 2 C - 1 O^{-}

New $N u - C_{β}$ bond at atom 4 — the far attack.
$C = C$ has moved inward (now between atoms 2 and 3) — the electrons rippled along.
$O^{-}$ at atom 1 — same destination as before; the H will land here.

This intermediate is an enolate — hold that thought; it must still turn into a normal carbonyl.

Step 6 — Tautomerise: the enolate becomes a carbonyl (and the C=C dies)

WHAT. The enolate $C = C - O^{-}$ picks up a proton on oxygen, giving an enol $C = C - O H$ , which flips to the more stable keto form $C - C = O$ . This flip is tautomerisation (see Enols and enolates).

WHY. A $C = O$ bond is far stronger than a $C = C$ bond ( $\approx 745$ vs $\approx 615 kJ/mol$ ). Nature pockets that energy: the enol collapses to the keto form, rebuilding $C = O$ and erasing the $C = C$ .

PICTURE. Watch the double bond hop from "between the carbons" (enol) to "carbon-to-oxygen" (keto); the H migrates from O to the α-carbon.

N u - C - α C = C - O^{-} H^{+} N u - C - α C = C - O H ⟶ N u - C - α C H - C = O

Middle (enol) — $C = C$ with an $O H$ ; unstable.
Right (keto) — $C = O$ restored, α-carbon now carries the new $H$ .
The old $C = C$ is gone.

Step 7 — Which path? Hard vs soft (the deciding rule)

WHAT. Same molecule, same two targets. The nucleophile decides.

WHY (HSAB). Atom 2 (carbonyl C) has a tight, localised δ⁺ → a hard electrophile. Atom 4 (β-C) has a spread-out, squishy δ⁺ → a soft electrophile. By the HSAB principle, like binds like: hard Nu → hard site (1,2); soft Nu → soft site (1,4).

PICTURE. Two roads from one fork: a small hard magnet snaps to the near end; a big soft magnet drifts to the far end.

Formula The fork, decided

Nucleophile	Nature	Adds	Product keeps
$R M g X$ , $R L i$ , $L i A l H_{4}$	hard, irreversible → kinetic	1,2	$C = C$ (allylic alcohol)
$R_{2} C uL i$ (cuprate), $R S^{-}$ , $R_{2} N H$ , enolates	soft, reversible → thermodynamic	1,4	$C = O$ (saturated carbonyl)
$C N^{-}$	low T → 1,2; warm → 1,4	both	temperature-switchable

Compare the twin substrate cases:

$C H_{3} M g B r$ + cyclohexenone → 1-methylcyclohex-2-en-1-ol (1,2; hard Grignard).
$(C H_{3})_{2} C uL i$ + cyclohexenone → 3-methylcyclohexan-1-one (1,4; soft cuprate).

Same enone, opposite answer — chosen purely by the metal.

Step 8 — Degenerate & edge cases (never get surprised)

WHAT / WHY / PICTURE, three ways the fork simplifies:

No conjugated C=C at all (plain aldehyde/ketone): atom 4 doesn't exist → only 1,2 is possible. The whole choice vanishes.
Blocked carbonyl (bulky groups crowd atom 2): even a moderately hard Nu is forced to the open β-carbon → 1,4 by sterics alone.
Reversible addition + heat (e.g. $C N^{-}$ , enolates): early on you may see 1,2, but given time and warmth the system un-adds and re-adds until it settles on the stabler 1,4. Kinetic first, thermodynamic last.

The one-picture summary

Everything above, on one canvas: one δ⁺ becomes two (resonance), the fork splits by hardness, and the two products differ in what double bond survives.

Recall Feynman retelling (plain words)

Start with a bar magnet that has one sticky end — that's a plain carbonyl, one hungry carbon. Now weld a second, springy segment onto it (the C=C). The stickiness leaks down the springy bit and pools at the far tip — so now there are two hungry spots: the near end and the far end. A tiny, fussy magnet (a Grignard) is fast and snaps onto the near end (1,2), leaving the springy double bond intact as an alcohol. A big, easygoing magnet (a cuprate, or a soft enolate) prefers the far end (1,4); when it lands there, the springy bond re-shuffles into the stronger carbonyl bond and disappears. Heat and patience let wishy-washy magnets like cyanide try both ends and eventually settle on the far end, because that far-end bond lasts longer. That "always choose the far end with a stabilised carbon magnet" move is the Michael addition — and its fingerprint is two carbonyls sitting 1,5 apart in the product.